Additive abelian group where x+x+x+x=0

[jjou]

(Problem 49 from practice GRE Math exam:) Up to isomorphism, how many additive abelian groups G of order 16 have the property that x + x + x + x = 0 for each x in G?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 5

The answer is (D) 3, but I don't understand what the problem is asking, really, and I don't know what strategy I should take for this problem. My first guess was to just construct some groups having this property, but I think, on the actual exam, this strategy would take too long.

I was able to construct two groups, but I can't figure out what the third is.

[tex]G_1=\{0, 1, x, x+1, x^2, x^2+1, x^2+x, x^2+x+1, x^3, x^3+1, x^3+x, x^3+x+1, x^3+x^2, x^3+x^2+1, x^3+x^2+x, x^3+x^2+x+1\}[/tex]

[tex]G_2=\{0, 1, 2, 3, x, x+1, x+2, x+3, 2x, 2x+1, 2x+2, 2x+3, 3x, 3x+1, 3x+2, 3x+3\}[/tex]

Does anyone know what the third group is or have a better method for solving this problem?
Any suggestions / insights would be appreciated!

 

[Dick]

Any finite abelian group can be expressed as the direct sum of cyclic groups whose orders are powers of primes. How many ways can you do that so the order is 16 and you still have 4x=0 for all x?

 

[jjou]

Okay, had to do some research to understand that. Is direct sum of cyclic groups the same as the direct product of groups?

If so, here are the 3 ways:

[tex]G_1=\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2[/tex]
[tex]G_2=\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_4[/tex]
[tex]G_3=\mathbb{Z}_4 \oplus \mathbb{Z}_4[/tex]

Here, each cyclic group has an order which divides 4, thus 4x=0 for every element. The order of the direct sum of the groups is the product of the orders of each group, which is 16 in each case.

Yes?

Thanks! :)

 

[Dick]

That's all there is to it.